Question: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $r \neq 0$. $x = \dfrac{-5r + 45}{3r - 6} \times \dfrac{7r - 14}{r^2 - 10r + 9} $
Answer: First factor the quadratic. $x = \dfrac{-5r + 45}{3r - 6} \times \dfrac{7r - 14}{(r - 9)(r - 1)} $ Then factor out any other terms. $x = \dfrac{-5(r - 9)}{3(r - 2)} \times \dfrac{7(r - 2)}{(r - 9)(r - 1)} $ Then multiply the two numerators and multiply the two denominators. $x = \dfrac{ -5(r - 9) \times 7(r - 2) } { 3(r - 2) \times (r - 9)(r - 1) } $ $x = \dfrac{ -35(r - 9)(r - 2)}{ 3(r - 2)(r - 9)(r - 1)} $ Notice that $(r - 2)$ and $(r - 9)$ appear in both the numerator and denominator so we can cancel them. $x = \dfrac{ -35\cancel{(r - 9)}(r - 2)}{ 3(r - 2)\cancel{(r - 9)}(r - 1)} $ We are dividing by $r - 9$ , so $r - 9 \neq 0$ Therefore, $r \neq 9$ $x = \dfrac{ -35\cancel{(r - 9)}\cancel{(r - 2)}}{ 3\cancel{(r - 2)}\cancel{(r - 9)}(r - 1)} $ We are dividing by $r - 2$ , so $r - 2 \neq 0$ Therefore, $r \neq 2$ $x = \dfrac{-35}{3(r - 1)} ; \space r \neq 9 ; \space r \neq 2 $